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My teacher taught me that the frequency representation of time limited signal will be band unlimited.

Also the frequency representation of time unlimited will be band limited.

i.e frequency domain representation of rectangular function will be sinc which is perfectly fine.Fourier Transform

Now,the question I have is if I have series of rectangular pulses

Rectangular pulses

Why don't I get band limited signal?

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  • \$\begingroup\$ @RogerRowland what? \$\endgroup\$ – nrb Jul 23 '16 at 6:47
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    \$\begingroup\$ If we assume that A implies B, the absence of A never implies the absence of B. That's basic logic. \$\endgroup\$ – Ignacio Vazquez-Abrams Jul 23 '16 at 6:56
  • \$\begingroup\$ A time unlimited signal can be band unlimited. For example, Fourier Series tells us that a periodic signal, eg square wave, may have harmonics to infinity. \$\endgroup\$ – Chu Jul 23 '16 at 7:47
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Many people mistake necessary and sufficient reasons. Or if they don't mistake them, they don't spell out the difference, thinking it's obvious.

A time-limited signal will be unlimited in the frequency domain.

It's necessary that a signal be time unlimited, for it to be able to have a limited band in the frequency domain, but that's still not sufficient.

We can construct the spectrum of the rectangular pulse train using the convolution rule. As the rectangular pulse train is a pulse, convolved with an infinite series of impulses, its spectrum will be the spectrum of the pulse, multiplied by the spectrum of an infinite series of impulses.

As the spectrum of an infinite series of impulses is itself an infinite series of impulses, the total spectrum will still go out to infinity.

If instead you had a time signal that was a rectangular pulse convolved with (say) a Gaussian pulse (which is limited in both time and in its spectrum), then the spectrum of that would indeed be frequency limited.

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  • \$\begingroup\$ A Gaussian pulse is not time or spectrum limited, although it does decay very fast in both. I'm not quite sure what a "Gaussian impulse" is, though? \$\endgroup\$ – Timo Jul 25 '16 at 8:23
  • \$\begingroup\$ I mean guassian pulse, an impulse is a delta of course. You are right theoretically of course, though wrong in practice, ie usefully. In order to get round what we really mean by DC, unlimited, infinity etc, it's often said the guassian pulse is 'compact' in both domains, thus fudging whether it goes to infinity or not. If you understand the difference, you know what I'm talking about, if you don't, then you don't need to know! It sounded like the OP didn't. If we look at how fast the power within a bandwidth approaches the total power, we can agree that 'compact' is a useful concept. \$\endgroup\$ – Neil_UK Jul 25 '16 at 8:27
  • \$\begingroup\$ I come from a math/physics background, where we usually do pay attention to such distinctions :) Of course the concept you're talking about is indeed very useful often in practice, and in maths you then typically embody these in concepts of limits and/or decay rates of functions. However, as the OP is referring to a mathematical theorem, the distinction is crucial here: the Gaussian pulse would be a counterexample to the (corrected) claim unless by band/time limited you meant strictly zero outside some interval. \$\endgroup\$ – Timo Jul 25 '16 at 9:47
  • \$\begingroup\$ By the way, "compact" is a very overloaded term when it comes to Fourier transforms/functional analysis, as for example "compactly supported" means exactly that a function is strictly zero outside some compact set. I'm not disputing the usefulness of the concept you point to, or trying to change engineering parlance if that is indeed common terminology, just pointing out that there's a bit of a need to be careful here. \$\endgroup\$ – Timo Jul 25 '16 at 9:52
  • \$\begingroup\$ @Timo I come from a maths/physics/engineering background. To engineers I'm a mathematician, to mathematicians I'm a physicist, and to physicists I'm only an engineer! Engineers only deal with practicalities, so I tend to play fast and loose with limits. An engineer can build a synthesiser that can generate an irrational frequency, whereas a mathematician can prove you cannot find a rational representation of an irrational number. The difference is how good is an approximation? Irreducible noise provides the delta below which the error of an approximation is 'good enough'. \$\endgroup\$ – Neil_UK Jul 25 '16 at 10:25
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The second sentence is wrong. The two true sentences are:

  1. The frequency representation of a time limited signal will be band unlimited.

  2. A band limited signal (in frequency) will be time unlimited.

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